(a) Substituting the line into the circle gives \[x^2+(2x-1)^2=2\quad\Longrightarrow\quad5x^2-4x-1=(5x+1)(x-1)=0.\] A is the upper intersection, so \(x=1,y=1\), giving \(A=(1,1)\).

(b) Here \(D=(1,0)\), and the right end of the circle is \(x=\sqrt2\). Rotating the region produces discs of squared radius \(y^2=2-x^2\): \[V=\pi\int_1^{\sqrt2}(2-x^2)\,dx=\pi\left[2x-\frac{x^3}{3}\right]_1^{\sqrt2}=\frac\pi3(4\sqrt2-5).\]

(c) The perimeter includes AD of length 1, the x-axis segment of length \(\sqrt2-1\), and the arc subtending \(\pi/4\) at the origin. Thus \[P=1+(\sqrt2-1)+\sqrt2\frac\pi4=\sqrt2\left(1+\frac\pi4\right).\]